PHP form doesn't send data to MySQL

Question

This is the shortest of two forms that wont send data to the database, but I have one that does work. When I press submit the form refreshes, clears the page and doesn't send the data to my table. I've included my HTML incase that is a problem.

include 'dbc.php';

$err = array();

if(@$_POST['doPersonal'] == 'Personal') 
{

This code filters harmful script code and escapes data of all POST data from the user submitted form.

foreach($_POST as $key => $value) {
    $data[$key] = filter($value);
}

Automatically collects the hostname or domain like example.com)

$host  = $_SERVER['HTTP_HOST'];
$host_upper = strtoupper($host);
$path   = rtrim(dirname($_SERVER['PHP_SELF']), '/\\');

if(empty($err)) {

SQL Insert

$sql_insert = "INSERT into `personal`
    (`sex`,`aux_citizen`,`birth_place`,`birth_country`,`children`)
    VALUES
    ('$data[sex]','$data[aux_citizen]','$data[birth_place]','$data[birth_country]',
    '$data[children]')";

mysql_query($sql_insert,$link) or die("Insertion Failed:" . mysql_error());

header("Location: thankyou.php");  
exit();
} 
}       
?>

HTML code for the form:

<html>
<head>
    <title>title</title>
    <meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1">
    <script language="JavaScript" type="text/javascript"
            src="file:///js/jquery.validate.js"></script>
    <tr>
    </tr>
    <script>
        $("#regForm").validate();
        })
        ;
    </script>
    <link href="styles.css" rel="stylesheet" type="text/css">
</head>
<body>
<table width="100%" border="0" cellspacing="0" cellpadding="5" class="main">
    <tr>
        <td colspan="3">&nbsp;</td>
    </tr>
    <tr>
        <td width="160" valign="top"><p>&nbsp;</p>

            <p>&nbsp; </p>

            <p>&nbsp;</p>

            <p>&nbsp;</p>

            <p>&nbsp;</p></td>
        <td width="732" valign="top"><p>
            <?php
            if (!empty($err)) {
                echo "<div class=\"msg\">";
                foreach ($err as $e) {
                    echo "* $e <br>";
                }
                echo "</div>";
            }
            ?>
            <br>

            <form action="personal.php" method="post" name="regForm" id="regForm">
                <table width="95%" border="0" cellpadding="3" cellspacing="3" class="forms">
                    <tr>
                        <td>Sex <font color="#CC0000">*</font></span></td>
                        <td><select name="sex" class="required" id="sex">
                            <option value="" selected></option>
                            <option value="M">Male</option>
                            <option value="F">Female</option>
                        </td>
                    </tr>
                    </select>

I wont put the rest of the HTML code as it will go on forever, but this is the submit function

<input name="doPersonal" type="submit" id="doPersonal" value="Submit">

Can anyone see the problem? It has worked once but now it won't Thanks in advance

Your code, specifically the HTML is severely malformed. You are also wide open to SQL injections, plus, you're suppressing errors. I suggest you read some good books about HTML, PHP and SQL. — Madara's Ghost
– Madara's Ghost, Commented Mar 9, 2012 at 11:50
@Truth Looks actually like it is being filtered against SQL injection: $data[$key] = filter($value); — Michael Berkowski
– Michael Berkowski, Commented Mar 9, 2012 at 11:57

Michael Berkowski · Accepted Answer · 2012-03-09 11:47:52Z

It looks like the name attribute and value attribute of your submit button are inconsistent with your check in the processor form:

// Submit's value should be 'Personal'
if(@$_POST['doPersonal'] == 'Personal') 


// But your value on the button is 'Submit'
<input name="doPersonal" type="submit" id="doPersonal" value="Submit">

// You can change this to:
if($_POST['doPersonal'] == 'Submit') 
// But see the improved version below...

Instead of using the @ to suppress errors, check for the presence of the array key before attempting to use it:

if (!empty($_POST['doPersonal']) && $_POST['doPersonal'] == 'Submit') {
   // Handle your form
}

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